I refer to this question, I asked recently in the physics board:


It doesn't make sense to type it again, so I link to it. Maybe the question can be answered here.


2 Answers 2


You may be forgetting that pressure also decreases with height (exponentially). Also, because $P=\rho R T$, $\frac{dP}{dT}=\rho R$ (that is, $c_p$ does not appear). But I digress in answering your question.

Let's break down why potential temperature increases with height. Let's start with the equation: $$\theta=T\left(\frac{P_0}{P}\right)^{\frac{R_d}{c_p}}\tag{1}$$ Now, if we want to ask why $\theta$ increases with height, let's rephrase $(1)$ as a function of height: $$\theta(z)=T(z)\left[\frac{P_0}{P(z)}\right]^{\frac{R_d}{c_p}}\tag{2}$$ Now if you follow my derivation for finding $$\frac{\partial \theta}{\partial z}=\frac{\theta}{T}\left(\frac{\partial T}{\partial z}+\frac{g}{c_p}\right)$$

You may note that the $\frac{\partial\theta}{\partial z}=0$ iff. $\frac{\partial T}{\partial z}=-\frac{g}{c_p}=-9.8 \textrm{ K km}^{-1}$ (dry adiabatic lapse rate). If $\frac{\partial T}{\partial z}>-\frac{g}{c_p}$ then $\frac{\partial \theta}{\partial z}>0$.

This does not mean that the parcel is not adiabatic, though. For a parcel to be truly adiabatic, $\frac{\partial \theta}{\partial t}+u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}+w\frac{\partial \theta}{\partial z}=0$ Therefore, the only time a certain location can be considered adiabatic and have $\frac{\partial \theta}{\partial z}=0$ is when there is no heating ($\frac{\partial \theta}{\partial t}=0$), and no horizontal advection of $\theta$ ($u\frac{\partial \theta}{\partial x}+v\frac{\partial \theta}{\partial y}=0$). Note, I did not exclude vertical advection of $\theta$, because that is implied in the condition that $\frac{\partial \theta}{\partial z}=0$.

I'll also take pause here and list a few diabatic processes, and why there may be some justification to ignore them for your mental image.

  • Radiation: with the exception of greenhouse gases, the atmosphere is mostly transparent.
  • Surface heating/ turbulent transport: more important near the surface (note that $\theta$ is a minimum at the surface).
  • Clouds/microphysical processes: discrete and are never everywhere all the time and moisture is a completely separate variable that varies all the time. There is a way around that (to some extent- the equivalent potential temperature, for latent heat exchange).
  • Chemical processes: See below about how I remember $\theta$ increases with height.

If you want a good reminder of why $\theta$ increases with height, it might be easier to remember that $sgn \left( \frac{\partial \theta}{\partial z}\right)$ is a good indicator of static stability. And thunderstorms need to stop at some point, even if that is the tropopause (where $\frac{\partial \theta}{\partial z}>0$ due to ozone heating).

  • $\begingroup$ While reading I found my problem: I confused active transport of a particular parcel from z to z+Δz with two parcels 1, 2, spatially being separated by Δz at the same time. In the former case, assuming adiabatic state change, θ will not change while T changes by -g/c_p*Δz. However, for the second case there is no limiting restriction regarding θ1 and θ2, since 2 could be energetically more "charged" as compared to 1. (of course I know, that a positive gradient of θ(z) yields a statically stable layer structure, while negative layered atmospheric structures are intrinsically unstable.) $\endgroup$
    – MichaelW
    Sep 9, 2020 at 20:14

If we'd be living in a dry atmosphere your reasoning is indeed correct. Air would rise adiabatically and air would loose about 9.8 °C/km (dry adiabatic lapse rate). This means constant potential temperature. However, Earths atmosphere isn't dry. As soon as a rising, moist air parcel reaches saturation, it will rise with a moist adiabatic lapse rate (6-7 °C/km). Potential temperature will rise (because if the parcel falls down again it will be warmer at the same level - see below). This causes clouds and sometimes rain and thus, reduces the overall water content in an air parcel.

If such a parcel is now going to fall again it will gain temperature according to the dry adiabatic lapse rate (because it lost water during ascension). Lets say initially a parcel starts rising with moist adiabatic lapse rate with temperature $T(t=0)$ and then falls down again and reaches the same level at $t=1$. You will then observe that $T(t=0)< T(t=1)$. Thus, potential temperature must have increased as the parcel was rising.

  • 1
    $\begingroup$ I think OP wants to know why potential temperature is not a constant as one goes up the troposphere. $\endgroup$
    – gansub
    Sep 8, 2020 at 14:56
  • $\begingroup$ And that is what I am answering. I'm not sure where my answer has flaws. If you have suggestions I'll gladly include them. $\endgroup$ Sep 8, 2020 at 15:36
  • $\begingroup$ yes but my "problem" was a wrong conclusion (see my comment above). $\endgroup$
    – MichaelW
    Sep 9, 2020 at 20:37

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